Sector of a Circle

5 Key excerpts on "Sector of a Circle"

• 

  eBook - PDF

  Elementary Geometry for College Students

  • Daniel C. Alexander, Geralyn M. Koeberlein(Authors)
  • 2019(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Thus, we make the following assumption about the measure of the area of a sector. Figure 8.48 POSTULATE 24 The ratio of the degree measure m of the arc (or central angle) of a sector to 360° is the same as the ratio of the area of the sector to the area of the circle; that is, area of sector area of circle 5 m 360 . Theorem 8.5.1 In a circle of radius length r, the area A of a sector whose arc has degree measure m is given by A 5 m 360 pr 2 . Theorem 8.5.1 follows directly from Postulate 24. The expression for the area of this sector can be written in the form m AB C 360 pr 2 . Segment of a Circle Area and Perimeter of a Segment Area of a Triangle with an Inscribed Circle 386 CHAPTER 8 ■ AREAS OF POLYGONS AND CIRCLES EXAMPLE 1 If muni2220O 5 100uni00B0 , find the area of the 100° sector shown in Figure 8.49. Use your calculator and round the answer to the nearest hundredth of a square inch. SOLUTION A 5 m 360 pr 2 becomes A 5 100 360 # p # 10 2 < 87.27 in 2 Figure 8.49 O A B 10 in. In applications with circles, we often seek exact answers for circumference and area; in such cases, we simply leave p in the result. For instance, in a circle of radius length 5 in., the exact circumference is 10p in. and the exact area is expressed as 25p in 2 . Because a sector is bounded by two radii and an arc, the perimeter of a sector is the sum of the lengths of the two radii and the length of its arc. In Example 2, we use the intercepted arc AB C and apply the formula, P sector 5 2r 1 / AB C . EXAMPLE 2 Given that muni2220O 5 100uni00B0, find the perimeter of the sector shown in Figure 8.49. Use the calculator value of p and round your answer to the nearest hundredth of an inch. SOLUTION Because r 5 10 and muni2220O 5 100uni00B0, / AB C 5 100 360 # 2 # p # 10 < 17.45 in. Now P sector 5 2r 1 / AB C becomes P sector 5 2(10) 1 17.45 < 37.45 in. Because a semicircle is one-half of a circle, a semicircular region corresponds to a central angle of 180°.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Mathematical Practices, Mathematics for Teachers

  Activities, Models, and Real-Life Examples

  • Ron Larson, Robyn Silbey(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Area of a Sector When the central angle of the sector is aº, the area A of the sector is A = ( a — 360 ) πr 2 . EXAMPLE 8 Finding the Area of a Sector EXAMPLE 9 Finding the Area of a Composite Figure Find the area of the portion of the basketball court. 19 ft 12 ft SOLUTION The figure consists of a rectangle and a semicircle. You can find the area of the entire figure by finding the sum of the areas of these two figures. Area of rectangle Area of semicircle A = (length)(width) A = ( a — 360 ) πr 2 = (19)(12) = ( 180 — 360 ) π(6) 2 = 228 = 18π ≈ 56.52 So, the area of the portion of the basketball court is about 228 + 56.52 = 284.52 square feet. The semicircle has a radius of 12 — 2 = 6 feet. Write formula for the area of a sector. Copyright 2014 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 464 Chapter 12 Circles and Circular Solids 1. Writing a Solution Key Write a solution key for the activity on page 458. Describe a rubric for grading a student’s work. 2. Grading the Activity In the activity on page 458, a student gave the answers below. Each question is worth 10 points. Assign a grade to each answer. For those that are incorrect, why do you think the student erred? Sample Student Work The area of the circle is twice the square of the radius. r r 2 r 2 2 3. Finding the Diameter and the Radius of a Circle a. Find the diameter b. Find the radius of the circle. of the circle. 3 m 14 in. 4. Finding the Diameter and the Radius of a Circle a. Find the diameter b. Find the radius of the circle. of the circle.

  Sign up to read

  Learn more about book
• 

  eBook - PDF

  Trigonometry

  • Cynthia Y. Young(Author)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  Round the answer to three significant digits. Solution Write the formula for area of a circular sector. A = 1 __ 2 r 2 θ r Substitute r = 2.2 centimeters and θ = 25° into the formula. A = 1 __ 2 (2.2 cm ) 2 (25°) Simplify. A = 60.5 cm 2 This is incorrect. What mistake was made? Conceptual In Exercises 79–82, determine whether each statement is true or false. 79. The length of an arc with central angle 45° in a unit circle is 45. 80. The length of an arc with central angle π __ 3 in a unit circle is π __ 3 . 81. If the radius of a circle doubles, then the arc length (associated with a fixed central angle) doubles. 82. If the radius of a circle doubles, then the area of the sector (associated with a fixed central angle) doubles. 83. If a smaller gear has radius r 1 , a larger gear has radius r 2 , and the smaller gear rotates θ° 1 , what is the degree measure of the angle the larger gear rotates? 84. If a circle with radius r 1 has an arc length s 1 associated with a particular central angle, write the formula for the area of the sector of the circle formed by that central angle, in terms of the radius and arc length. Challenge For Exercises 85–88, refer to the following: You may think that a baseball field is a circular sector, but it is not. If it were, the distances from home plate to left field, center field, and right field would all be the same (the radius). Where the infield dirt meets the outfield grass and along the fence in the outfield are arc lengths associated with a circle of radius 95 feet and with a vertex located at the pitcher’s mound (not home plate). (Source: Karp, Cell and Molecular Biology, 8e, Figure 13.5, p. 517.) Infield / Outfield Grass Line: 95-ft radius from front of pitching rubber Infield Second base First base 13-ft radius Third base 13-ft radius Home plate 13-ft radius Foul line Foul line 90 ft between bases Pitching Mound 9-ft Radius 85.

  Sign up to read

  Learn more about book
• 

  eBook - ePub

  Plain Plane Geometry

  • Amol Sasane(Author)
  • 2015(Publication Date)
  • WSPC

    (Publisher)

  Chapter 5

  Circles

  We had defined the terms circle, its center and its radius in Section 1.6 . In this chapter, we will study circles. We begin with some terminology. Recall that a circle C (O, r ) with center O and radius r > 0 is the set of points P in the plane such that OP = r . The set of points P in the plane such that OP < r is called the interior of the circle, while the set of points P for which OP > r is called the exterior of the circle. Circles having the same center are said to be concentric .

  A line segment joining any two points on the circle is called a chord .

  A diameter is a special type of chord: it is one which passes through the center of the circle. Clearly the length of any diameter of a circle of radius r is equal to 2r . A diameter divides the circle into two semicircles .

  Theorem 5.1 . The perpendicular from the center of a circle to a chord bisects the chord .

  Proof . Let OM be the perpendicular dropped from the center O of the circle C (O, r ) to its chord AB . In the right triangles ΔOAM and ΔOBM , we have OA = OB = r , and the side OM is common. By the RHS Congruency Rule, ΔOAM ≃ ΔOBM , giving AM = MB , as wanted.

  Theorem 5.2 . The line joining the center of a circle to the midpoint of a chord is perpendicular to the chord .

  Proof . Let M be the midpoint of the chord AB of the circle C (O, r ). In the two triangles ΔOAM and ΔOBM , we have OA = OB = r , the side OM is common, and AM = BM . By the SSS Congruency Rule, we have ΔOAM ≃ ΔOBM . Hence we obtain ∠OMA = ∠OMB , and being supplementary, they must each equal 90°.

  Corollary 5.1 . The perpendicular bisectors of two chords of a circle contain its center .

  We know that two distinct points determine a unique line passing through them. We can now ask:

  Sign up to read

  Learn more about book
• 

  eBook - ePub

  Journey into Mathematics

  An Introduction to Proofs

  • Joseph J. Rotman(Author)
  • 2013(Publication Date)
  • Dover Publications

    (Publisher)

  Section 2 . It is impossible to compute the area of a circle on the diameter as the linear unit without trespassing upon the area outside of the circle to the extent of including one- fifth more area than is contained within the circle’s circumference, because the square on the diameter produces the side of a square which equals nine when the arc of ninety degrees equals eight. But taking the quadrant of the circle’s circumference for the linear unit, we fulfill the requirements of both quadrature and rectification of the circle’s circumference. Furthermore, it has revealed the ratio of the chord and arc of ninety degrees, which is as seven to eight, and also the ratio of the diagonal and one side of a square which is as ten to seven, disclosing the fourth important fact, that the ratio of the diameter and circumference is as five-fourths to four; and because of these facts and the further fact that the rule in present use fails to work both ways mathematically, it should be discarded as wholly wanting and misleading in its practical applications.

  Section 3. In further proof of the value of the author’s proposed contribution to education, and offered as a gift to the State of Indiana, is the fact of his solutions of the trisection of the angle, duplication of the cube and quadrature of the circle having been already accepted as contributions to science by the American Mathematical Monthly, the leading exponent of mathematical thought in this country. And be it remembered that these noted problems had been long since given up by scientific bodies as unsolvable mysteries and above man’s ability to comprehend.

  I display the bill for two reasons. The first, of course, is that one ought to see what one is criticizing. Second, we have here a prime example of how not to write mathematics. I will try to decipher the beginning of Section 1, but it is so poorly written that I can only guess at its meaning, if any. Given a circle of radius r , its “circular area” is πr 2 ; “quadrant” means one-fourth, so that a “quadrant of a circumference” seems to mean 2πr /4 = πr /2; “square” seems to mean the algebraic operation of multiplying a number by itself, whereas “equilateral rectangle” seems to mean the geometric figure we call a square (otherwise why not call it a square?). Thus, the text of the bill seems to say that πr 2 /(πr /2)2 = r 2 /r 2 ; that is, the bill wants to legislate the value of π to be 4 (it appears there are at least three other values for π

  Sign up to read

  Learn more about book